The Notebook

In this post, I show that we have infinitely many prime numbers. To do so, I use a proof from 300 BC which
provided by Greek Mathematician. The one that people usually call as “Father of Geometry” or **Euclid** of Alexandria.

Here we break into proof:

Let's write prime numbers in an ascending manner, ** p_1 = 2, p_2 = 3, p_3 = 5, p_4 = 7**

When we check the numbers which are in a pattern of: ** p_1.p_2 + 1 = 2.3 + 1 = 7** and ** p_1.p_2.p_3 + 1 = 2.3.5 + 1 = 31**
and ** p_1.p_2.p_3.p_4 + 1 = 2.3.5.7 + 1 = 211** we oftenly come up with a prime number. But how can we be
sure about primality of number ** A = p_1.p_2 \dots p_n + 1** for all values of **n**?

Here we use contradiction technique to check: We intuitively think that **A** is a prime number, since it empirically is.
Now we set up an equation which assumes that **A** is a non-prime number and try to show that this equation would not hold -hence contradicts.

Following equation assumes that **A** is a divisor of some number. We clearly know that, every number -except 1- is at least multiple
of one prime number. For this reason, let's say that ** p_r** is a divisor of **A**.

** A = p_1.p_2.p_3 \dots p_r \dots p_n + 1 \rightarrow \frac{A}{p_r} = p_1.p_2.p_3 \dots p_{r-1}.p_{r+1} \dots p_n + \frac{1}{p_r}**.
Consequently we have ** \frac{A}{p_r} - p_1.p_2.p_3 \dots p_{r-1}.p_{r+1} \dots p_n = \frac{1}{p_r}**

Let's concentrate on last equation. ** \frac{A}{p_r}** has to be an integer, since we assume **A** is a
multiple of ** p_r**. Second term ** p_1.p_2.p_3 \dots p_{r-1}.p_{r+1} \dots p_n** is also integer, nevertheless
** \frac{1}{p_r}** is not. So, our non-primality assumption of **A** was wrong and this wrong assumption proves us that,
**A** would always be prime for any value of n.